Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Univariate continuous distributions ▷ Normal distribution ▷ Cumulative distribution function without error function

Theorem: Let $X$ be a random variable following a normal distribution:

\[\label{eq:norm} X \sim \mathcal{N}(\mu, \sigma^2) \; .\]

Then, the cumulative distribution function of $X$ can be expressed as

\[\label{eq:norm-cdf} F_X(x) = \Phi_{\mu,\sigma}(x) = \varphi\left( \frac{x-\mu}{\sigma} \right) \cdot \sum_{i=1}^{\infty} \frac{\left( \frac{x-\mu}{\sigma} \right)^{2i-1}}{(2i-1)!!} + \frac{1}{2}\]

where $\varphi(x)$ is the probability density function of the standard normal distribution and $n!!$ is a double factorial.

Proof:

1) First, consider the standard normal distribution $\mathcal{N}(0, 1)$ which has the probability density function

\[\label{eq:snorm-pdf} \varphi(x) = \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{1}{2} x^2} \; .\]

Let $T(x)$ be the indefinite integral of this function. It can be obtained using infinitely repeated integration by parts as follows:

\[\label{eq:snorm-pdf-ii-s1} \begin{split} T(x) &= \int \varphi(x) \, \mathrm{d}x \\ &= \int \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{1}{2} x^2} \, \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi}} \int 1 \cdot e^{-\frac{1}{2} x^2} \, \mathrm{d}x \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \left[ x \cdot e^{-\frac{1}{2} x^2} + \int x^2 \cdot e^{-\frac{1}{2} x^2} \, \mathrm{d}x \right] \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \left[ x \cdot e^{-\frac{1}{2} x^2} + \left[ \frac{1}{3} x^3 \cdot e^{-\frac{1}{2} x^2} + \int \frac{1}{3} x^4 \cdot e^{-\frac{1}{2} x^2} \, \mathrm{d}x \right] \right] \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \left[ x \cdot e^{-\frac{1}{2} x^2} + \left[ \frac{1}{3} x^3 \cdot e^{-\frac{1}{2} x^2} + \left[ \frac{1}{15} x^5 \cdot e^{-\frac{1}{2} x^2} + \int \frac{1}{15} x^6 \cdot e^{-\frac{1}{2} x^2} \, \mathrm{d}x \right] \right] \right] \\ &= \ldots \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \left[ \sum_{i=1}^{n} \left( \frac{x^{2i-1}}{(2i-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) + \int \left( \frac{x^{2n}}{(2n-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) \, \mathrm{d}x \right] \\ &= \frac{1}{\sqrt{2 \pi}} \cdot \left[ \sum_{i=1}^{\infty} \left( \frac{x^{2i-1}}{(2i-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) + \lim_{n \to \infty} \int \left( \frac{x^{2n}}{(2n-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) \, \mathrm{d}x \right] \; . \end{split}\]

Since $(2n-1)!!$ grows faster than $x^{2n}$, it holds that

\[\label{eq:int-const} \frac{1}{\sqrt{2 \pi}} \cdot \lim_{n \to \infty} \int \left( \frac{x^{2n}}{(2n-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) \, \mathrm{d}x = \int 0 \, \mathrm{d}x = c\]

for constant $c$, such that the indefinite integral becomes

\[\label{eq:snorm-pdf-ii-s2} \begin{split} T(x) &= \frac{1}{\sqrt{2 \pi}} \cdot \sum_{i=1}^{\infty} \left( \frac{x^{2i-1}}{(2i-1)!!} \cdot e^{-\frac{1}{2} x^2} \right) + c \\ &= \frac{1}{\sqrt{2 \pi}} \cdot e^{-\frac{1}{2} x^2} \cdot \sum_{i=1}^{\infty} \frac{x^{2i-1}}{(2i-1)!!} + c \\ &\overset{\eqref{eq:snorm-pdf}}{=} \varphi(x) \cdot \sum_{i=1}^{\infty} \frac{x^{2i-1}}{(2i-1)!!} + c \; . \end{split}\]

2) Next, let $\Phi(x)$ be the cumulative distribution function of the standard normal distribution:

\[\label{eq:snorm-cdf} \Phi(x) = \int_{-\infty}^x \varphi(x) \, \mathrm{d}x \; .\]

It can be obtained by matching $T(0)$ to $\Phi(0)$ which is $1/2$, because the standard normal distribution is symmetric around zero:

\[\label{eq:snorm-cdf-c} \begin{split} T(0) = \varphi(0) \cdot \sum_{i=1}^{\infty} \frac{0^{2i-1}}{(2i-1)!!} + c &= \frac{1}{2} = \Phi(0) \\ \Leftrightarrow c &= \frac{1}{2} \\ \Rightarrow \Phi(x) = \varphi(x) \cdot \sum_{i=1}^{\infty} \frac{x^{2i-1}}{(2i-1)!!} + \frac{1}{2} \! &\; . \end{split}\]

3) Finally, the cumulative distribution functions of the standard normal distribution and the general normal distribution are related to each other as

\[\label{eq:norm-snorm-cdf} \Phi_{\mu,\sigma}(x) = \Phi\left( \frac{x-\mu}{\sigma} \right) \; .\]

Combining \eqref{eq:norm-snorm-cdf} with \eqref{eq:snorm-cdf-c}, we have:

\[\label{eq:norm-cdf-qed} \Phi_{\mu,\sigma}(x) = \varphi\left( \frac{x-\mu}{\sigma} \right) \cdot \sum_{i=1}^{\infty} \frac{\left( \frac{x-\mu}{\sigma} \right)^{2i-1}}{(2i-1)!!} + \frac{1}{2} \; .\]
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Metadata: ID: P86 | shortcut: norm-cdfwerf | author: JoramSoch | date: 2020-03-20, 04:26.